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mihalych1998 [28]
3 years ago
12

Need help to answer this problem

Mathematics
1 answer:
pashok25 [27]3 years ago
3 0

Answer:

Step-by-step explanation:

9.1 h + 11.6 t = 9.98(60)

9.1 h + 11.6 t = 598.8 ........ <em>(1)</em>

h + t = 60 ........ <em>(2)</em>

h = 60 - t

9.1(60 - t) + 11.6 t = 598.8

546 - 9.1 t + 11.6 t = 598.8

2.5 t = 52.8

<em>t</em> = 52.8 ÷ 2.5 = <em>21.12</em><em> </em><em>≈</em><em> 21</em> (lb.)

<em>h</em> = 60 - 21.12 = <em>38.88 ≈</em><em> 39</em> (lb.)

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All angles would have a sum of 180 degrees, so 180-140 is 40. And if you divide 40 by 2, the two side angles will be 20 degrees.
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Read the problems and and write the word that indicates an equation or inequality and decide whether an equation or inequality i
Gnoma [55]

Answer:

Word/phrase : Equation

Equation:4x + 2 = $30

He can afford to buy 7 books without spending more than his gift certificate amount

Step-by-step explanation:

Step 1

We find the equation

Brett has a $30 online gift certificate. He plans to buy as many books as he can. The cost of each book is $4. There is also a single shipping charge of $2.

Hence,

$4 × x + $2 = $30

Where x = number of books that he can buy

4x + 2 = $30

Step 2

We solve for x

4x + 2 = 30

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x = 28/4

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Hence, Brett can buy 7 books

Equation: 4x + 2 = $30

Word/phrase : Equation

He can afford to buy 7 books without spending more than his gift certificate amount

8 0
3 years ago
The mean points obtained in an aptitude examination is 159 points with a standard deviation of 13 points. What is the probabilit
Korolek [52]

Answer:

0.4514 = 45.14% probability that the mean of the sample would differ from the population mean by less than 1 point if 60 exams are sampled

Step-by-step explanation:

To solve this question, we have to understand the normal probability distribution and the central limit theorem.

Normal probability distribution:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

Central limit theorem:

The Central Limit Theorem estabilishes that, for a random variable X, with mean \mu and standard deviation \sigma, a large sample size can be approximated to a normal distribution with mean \mu and standard deviation s = \frac{\sigma}{\sqrt{n}}

In this problem, we have that:

\mu = 159, \sigma = 13, n = 60, s = \frac{13}{\sqrt{60}} = 1.68

What is the probability that the mean of the sample would differ from the population mean by less than 1 point if 60 exams are sampled?

This is the pvalue of Z when X = 159+1 = 160 subtracted by the pvalue of Z when X = 159-1 = 158. So

X = 160

Z = \frac{X - \mu}{\sigma}

By the Central Limit Theorem

Z = \frac{X - \mu}{s}

Z = \frac{160 - 159}{1.68}

Z = 0.6

Z = 0.6 has a pvalue of 0.7257

X = 150

Z = \frac{X - \mu}{s}

Z = \frac{158 - 159}{1.68}

Z = -0.6

Z = -0.6 has a pvalue of 0.2743

0.7257 - 0.2743 = 0.4514

0.4514 = 45.14% probability that the mean of the sample would differ from the population mean by less than 1 point if 60 exams are sampled

7 0
3 years ago
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